Worked examples — Ohm's law — microscopic origin, resistivity

This page builds on the parent note and touches Electric current and current density, Drude model of conduction, Temperature dependence of resistance, Joule heating — power dissipation, and Resistors in series and parallel.

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The scenario matrix

Before solving anything, let's list every distinct kind of problem this topic can ask. Each worked example below is tagged with the cell it fills.

Cell	What makes it distinct	Covered by
A. Basic forward	Given geometry + $\rho$, find $R$	Ex 1
B. Basic inverse	Given $R$, geometry, find $\rho$ (material ID)	Ex 2
C. Micro chain	From $E$ (or $V$) → $v_d$ → $J$ → $I$	Ex 3
D. Scaling / ratios	"Double $L$", "halve radius" — no numbers needed	Ex 4
E. Limiting / degenerate	$\tau \to 0$, $n\to 0$, $A\to 0$, insulator/superconductor	Ex 5
F. Temperature case	$\tau$ changes with $T$; sign of the change	Ex 6
G. Real-world word problem	Household wire heating up	Ex 7
H. Exam twist	Two wires, same material, series/parallel combo	Ex 8

Signs never turn negative here (all of $n,e^2,\tau ,m,L,A$ are positive), so the "quadrant" cases of trig-style problems become limiting cases instead: what happens as each quantity heads to $0$ or $\infty$. Cell E hunts those down deliberately.

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Setup — the constants we reuse

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Cell A — Basic forward

Step 1 — Convert the radius to metres and get the area. $r=0.2\text{mm}=0.2\times 10^{-3}\text{m}=2\times 10^{-4}\text{m}$ $A=\pi r^2=\pi (2\times 10^{-4}{)}^2=\pi \times 4\times 10^{-8}\approx 1.257\times 10^{-7}{\text{m}}^2$ Why this step? $R=\rho L/A$ needs area, and the wire is round, so $A=\pi r^2$. Millimetres must become metres or the powers of ten go wrong.

Step 2 — Plug into $R=\rho L/A$. $R=\frac{(5.6\times 10^{-8})(3)}{1.257\times 10^{-7}}\approx 1.34\Omega$ Why this step? This is the master geometry formula from the parent note.

Verify: Units: $\frac{\Omega \text{m}\cdot \text{m}}{{\text{m}}^2}=\Omega$. ✓ Order of magnitude: numerator $\sim 10^{-7}$, denominator $\sim 10^{-7}$, so $R\sim 1\Omega$. Matches forecast "a few ohms." ✓

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Cell B — Basic inverse (identify the material)

Step 1 — Area of the square cross-section. $A=(2\times 10^{-3})(2\times 10^{-3})=4\times 10^{-6}{\text{m}}^2$ Why this step? Cross-section is a square, so $A={\text{side}}^2$; still convert mm→m first.

Step 2 — Invert $R=\rho L/A$ to solve for $\rho$. $\rho =\frac{RA}{L}=\frac{(2.125\times 10^{-3})(4\times 10^{-6})}{0.5}=1.7\times 10^{-8}\Omega \text{m}$ Why this step? We're given $R$ and asked for $\rho$, so rearrange — multiply both sides by $A/L$.

Verify: $1.7\times 10^{-8}$ is exactly copper's resistivity → the rod is copper. Units: $\frac{\Omega \cdot {\text{m}}^2}{\text{m}}=\Omega \text{m}$. ✓

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Cell C — The full microscopic chain

Step 1 — Drift speed from $v_d=eE\tau /m$. $v_d=\frac{(1.6\times 10^{-19})(0.1)(8\times 10^{-15})}{9.1\times 10^{-31}}\approx 1.41\times 10^{-4}\text{m/s}$ Why this step? The parent's Step 3 result: field pushes for one free-flight $\tau$ before a collision resets it. Look at the figure — the jagged path is thermal wandering; the tiny rightward slope is $v_d$.

Step 2 — Current density $J=ne{v}_d$. $J=(1.8\times 10^{29})(1.6\times 10^{-19})(1.406\times 10^{-4})\approx 4.05\times 10^6{\text{A/m}}^2$ Why this step? Count how much charge sweeps through per second per unit area — the parent's Step 4.

Step 3 — Current $I=JA$. $I=(4.05\times 10^6)(10^{-6})\approx 4.05\text{A}$ Why this step? Current is current-density spread over the whole cross-section.

Verify: Cross-check via $\sigma$: $\sigma =n{e}^2\tau /m=(1.8\times 10^{29})(1.6\times 10^{-19}{)}^2(8\times 10^{-15})/(9.1\times 10^{-31})\approx 4.05\times 10^7\text{S/m}$, so $J=\sigma E=4.05\times 10^6$. ✓ Same answer two ways. Drift speed $\sim 10^{-4}\text{m/s}$ — a crawl, as forecast. ✓

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Cell D — Scaling and ratios (no numbers)

Step 1 — (a) Double length, area fixed. $R=\rho \frac{L}{A}\Rightarrow R\propto L\Rightarrow R_a=2R_0$ Why this step? $\rho$ and $A$ unchanged, so $R$ scales straight with $L$ — twice the collisions in series.

Step 2 — (b) Halve radius, length fixed. $A=\pi r^2\Rightarrow r\to r/2⟹A\to A/4⟹R_b=\frac{R_0}{1/4}=4R_0$ Why this step? $R\propto 1/A$ and area goes as $r^2$, so halving $r$ quarters $A$ and quadruples $R$.

Step 3 — (c) Redraw to double length, same volume. Volume $V=AL$ is fixed. If $L\to 2L$ then $A\to A/2$. So $R_c=\rho \frac{2L}{A/2}=4\rho \frac{L}{A}=4R_0$ Why this step? Stretching a fixed lump of metal makes it both longer and thinner. Both effects push $R$ up, giving a factor $2\times 2=4$.

Verify: Ranking: (a) $2\times$, (b) $4\times$, (c) $4\times$. Thinning dominates. A general redraw rule: $R\propto L^2$ at fixed volume — double length → $4\times$. ✓

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Cell E — Limiting and degenerate cases

Step 1 — (i) $\tau \to \infty$. $\rho =\frac{m}{n{e}^2\tau }\to \frac{m}{n{e}^2\cdot \infty }=0\Rightarrow R\to 0$ Why this step? No collisions means the brake is gone — electrons drift freely, resistivity vanishes. This is the idealised superconductor limit.

Step 2 — (ii) $n\to 0$. $\rho =\frac{m}{n{e}^2\tau }\to \frac{m}{0}=\infty \Rightarrow R\to \infty$ Why this step? No free carriers, no current possible — a perfect insulator. This is why $n$ (not $\tau$) is the switch between metal and insulator (see Semiconductors vs metals).

Step 3 — (iii) $A\to 0$. $R=\rho \frac{L}{A}\to \rho \frac{L}{0}=\infty$ Why this step? Squeeze all the current through a vanishing cross-section → infinite resistance. Physically this is where a thin filament overheats and blows (link to Joule heating in Ex 7).

Step 4 — (iv) $L\to 0$. $R=\rho \frac{L}{A}\to \rho \frac{0}{A}=0$ Why this step? No length to travel → no accumulated collisions → zero resistance. A wide, flat slab is nearly a short.

Verify: Two ways to reach $R=0$ (perfect crystal, or zero length), two ways to reach $R=\infty$ (no carriers, or zero area). The material limits ($\tau ,n$) live in $\rho$; the shape limits ($A,L$) live in the geometry factor. ✓

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Cell F — Temperature case

Step 1 — Write $R$'s dependence on $\tau$. $R=\rho \frac{L}{A}=\frac{m}{n{e}^2\tau }\cdot \frac{L}{A}\Rightarrow R\propto \frac{1}{\tau }$ Why this step? Geometry and $n$ are held fixed, so only $\tau$ moves — and it sits in the denominator.

Step 2 — Take the ratio. $\frac{R_{\text{hot}}}{R_{\text{cold}}}=\frac{{\tau }_{\text{cold}}}{{\tau }_{\text{hot}}}=\frac{2.5\times 10^{-14}}{2.0\times 10^{-14}}=1.25$ Why this step? Ratios kill all the shared constants ($m,n,e,L,A$), leaving only the two $\tau$'s.

Verify: $R$ rises by 25%. Sign check: smaller $\tau$ → larger $1/\tau$ → larger $R$. Metals get more resistive when hot, exactly as Temperature dependence of resistance states. ✓

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Cell G — Real-world word problem

Step 1 — (a) Resistance from geometry. $A=0.3{\text{mm}}^2=0.3\times 10^{-6}{\text{m}}^2$ $R=\rho \frac{L}{A}=\frac{(1.1\times 10^{-6})(1.5)}{0.3\times 10^{-6}}=5.5\Omega$ Why this step? We need $R$ before any current or power — it's the object's fixed property.

Step 2 — (b) Current from Ohm's law. $I=\frac{V}{R}=\frac{230}{5.5}\approx 41.8\text{A}$ Why this step? Macroscopic $V=IR$, rearranged. (A real toaster uses a longer/thinner coil to pull less current; this simplified one is deliberately hungry.)

Step 3 — (c) Power from Joule heating. $P=VI=(230)(41.8)\approx 9.6\times 10^3\text{W}$ Why this step? Power = voltage × current — every joule of electrical energy becomes heat in the element (see Joule heating — power dissipation).

Verify: Cross-check $P=V^2/R=230^2/5.5=52900/5.5\approx 9.6\times 10^3\text{W}$. ✓ Same both ways. Units: $\text{V}\cdot \text{A}=\text{W}$. ✓

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Cell H — Exam twist (combination)

Step 1 — Get $R_2$ from the radius change. Twice the radius → $A_2=\pi (2r{)}^2=4\pi r^2=4A_1$. Since $R\propto 1/A$ (same $\rho$, $L$): $R_2=\frac{R_1}{4}=\frac{3}{4}=0.75\Omega$ Why this step? Only the area differs; area scales as radius squared, so $R$ drops by 4.

Step 2 — (a) Series. $R_{\text{series}}=R_1+R_2=3+0.75=3.75\Omega$ Why this step? In series the current threads through both in turn, so resistances add (parent note: "collisions in series").

Step 3 — (b) Parallel. $\frac{1}{R_{\text{par}}}=\frac{1}{3}+\frac{1}{0.75}=0.3333+1.3333=1.6667{\Omega }^{-1}$ $R_{\text{par}}=\frac{1}{1.6667}=0.6\Omega$ Why this step? Parallel paths add conductances; more parallel routes → lower total resistance (see Resistors in series and parallel).

Verify: Sanity rule: parallel total ($0.6\Omega$) is less than the smallest branch ($0.75\Omega$). ✓ Series total ($3.75\Omega$) is more than the largest branch ($3\Omega$). ✓ Both consistency checks pass; and $0.6<1$, confirming the forecast.

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Active Recall

Recall Which cell needs

$A=\pi r^2$ vs $A={\text{side}}^2$? Round wires use $A=\pi r^2$ (Ex 1, 3, 8); square/rectangular cross-sections use side×side (Ex 2).

Recall Redrawing a wire to double length at fixed volume multiplies

$R$ by… $4$ — because $R\propto L^2$ at constant volume (Ex 4c).

Recall Which quantity flips metal↔insulator,

$\tau$ or $n$? $n$: $n\to 0$ gives $\rho \to \infty$ (insulator); $\tau$ only tunes how good a conductor a metal is (Ex 5).

Recall Hotter metal:

$\tau$ does what to $R$? $\tau$ falls, and since $R\propto 1/\tau$, $R$ rises (Ex 6).

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Connections

• Ohm's law — microscopic origin, resistivity (index 1.8.16) (parent)
• Electric current and current density
• Drude model of conduction
• Resistors in series and parallel
• Temperature dependence of resistance
• Joule heating — power dissipation
• Semiconductors vs metals